Physics-
General
Easy

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T subscript 2 end subscript and T subscript 1 end subscript left parenthesis T subscript 2 end subscript greater than T subscript 1 end subscript right parenthesis. The rate of heat transfer through the slab, in a steady state is open parentheses fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close parentheses f, with f equals to

  1. 1    
  2. 1/3    
  3. 2/3    
  4. 1/3    

The correct answer is: 1/3


    Let the temperature of common interface be T ℃. Rate of heat flow
    H equals fraction numerator Q over denominator t end fraction equals blank fraction numerator K A increment T over denominator l end fraction
    therefore H subscript 1 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 1 end subscript=fraction numerator 2 K A left parenthesis T minus T subscript 1 right parenthesis end subscript over denominator 4 x end fraction
    And H subscript 2 end subscript equals open parentheses fraction numerator Q over denominator t end fraction close parentheses subscript 2 end subscript=fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    In steady state, the rate of heat flow should be same in whole system i e comma
    H subscript 1 end subscript equals blank H subscript 2 end subscript
    rightwards double arrow fraction numerator 2 K A left parenthesis T minus T subscript 1 end subscript right parenthesis over denominator 4 x end fraction = fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    rightwards double arrow fraction numerator T minus T subscript 1 end subscript over denominator 2 end fraction equals blank T subscript 2 end subscript minus T
    rightwards double arrow T minus T subscript 1 end subscript equals blank 2 T subscript 2 end subscript minus 2 T
    rightwards double arrow T equals fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction (i)
    Hence, heat flow from composite slab is
    H equals blank fraction numerator K A left parenthesis T subscript 2 end subscript minus T right parenthesis over denominator x end fraction
    = fraction numerator K A over denominator x end fraction open parentheses T subscript 2 end subscript minus fraction numerator 2 T subscript 2 end subscript plus T subscript 1 end subscript over denominator 3 end fraction close parentheses equals blank fraction numerator K A over denominator 3 x end fraction blank left parenthesis T subscript 2 end subscript minus T subscript 1 end subscript right parenthesis (ii)
    Accordingly, H equals open square brackets fraction numerator A open parentheses T subscript 2 end subscript minus T subscript 1 end subscript close parentheses K over denominator x end fraction close square brackets f (iii)
    By comparing Eqs. (ii) and (iii), we get
    rightwards double arrow f equals blank fraction numerator 1 over denominator 3 end fraction

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